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Question Number 92137 by hore last updated on 05/May/20
∫2x1+x
Commented by john santu last updated on 05/May/20
∫2xdxx+1=∫2(x+1)−2x+1dx=∫2−2x+1dx=2x−2ln∣x−1∣+c
Commented by Rio Michael last updated on 07/May/20
2x1+x=2−2x+1[canuselongdivision]∫2x1+xdx=∫(2−2x+1)dx=2x−2ln(x+1)+C
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